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Math Help - Limit

  1. #1
    MHF Contributor alexmahone's Avatar
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    Limit

    Find lim_{x\to\infty} x[(1+\frac{1}{x})^x-e] [Suggestion: Let u=\frac{1}{x}, and take the limit as u\to0.]
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  2. #2
    MHF Contributor chisigma's Avatar
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    Setting x=\frac{1}{u} and applying several times l'Hopital's rule is ...

    \displaystyle \lim_{u \rightarrow 0} \frac{(1+u)^{\frac{1}{u}} - e}{u}= \lim_{n \rightarrow 0} \frac{(1+u)^{\frac{1}{u}}}{1+u}\ \frac {u- \ln (1+u)}{u^{2}} = e\ \lim_{u \rightarrow 0} \frac{1- \frac{1}{1+u}}{2u} =

    \displaystyle = e \lim_{u \rightarrow 0} \frac{1}{2\ (1+u)^{2}} = \frac{e}{2}

    Kind regards

    \chi \sigma
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